//将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 
//
// 
//
// 示例 1： 
//
// 
//输入：l1 = [1,2,4], l2 = [1,3,4]
//输出：[1,1,2,3,4,4]
// 
//
// 示例 2： 
//
// 
//输入：l1 = [], l2 = []
//输出：[]
// 
//
// 示例 3： 
//
// 
//输入：l1 = [], l2 = [0]
//输出：[0]
// 
//
// 
//
// 提示： 
//
// 
// 两个链表的节点数目范围是 [0, 50] 
// -100 <= Node.val <= 100 
// l1 和 l2 均按 非递减顺序 排列 
// 
// Related Topics 递归 链表 
// 👍 2294 👎 0


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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {

        if (list1 == null && list2 == null){
            return null;
        }
        if (list1 == null){
            return list2;
        }
        if (list2 == null){
            return list1;
        }

        return mergeTwoNode(list1, list2);



    }


    public ListNode mergeTwoNode(ListNode nodeBig, ListNode nodeSmall){

        if (nodeBig.next == null && nodeSmall.next == null){
            if (nodeBig.val > nodeSmall.val){
                nodeSmall.next = nodeBig;
                return nodeSmall;
            }else {
                nodeBig.next = nodeSmall;
                return nodeBig;
            }
        }

        if (nodeBig.next != null && nodeSmall.next == null){
            if (nodeBig.val > nodeSmall.val){
                nodeSmall.next = nodeBig;
                return nodeSmall;
            }else {
                nodeBig.next = mergeTwoNode(nodeBig.next,nodeSmall);
                return nodeBig;
            }

        }

        if (nodeBig.next == null && nodeSmall.next != null){
            if (nodeSmall.val > nodeBig.val){
                nodeBig.next = nodeSmall;
                return nodeBig;
            }else {
                nodeSmall.next = mergeTwoNode(nodeSmall.next,nodeBig);
                return nodeSmall;
            }

        }
        if (nodeBig.val > nodeSmall.val){
            nodeSmall.next = mergeTwoNode(nodeSmall.next,nodeBig);
            return nodeSmall;
        }else {
            nodeBig.next = mergeTwoNode(nodeSmall,nodeBig.next);
            return nodeBig;
        }



    }
}
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